\(\left(a+b+c\right)\left(a^2+b^2+c^2\right)=a^3+ab^2+b^3+bc^2+c^3+ca^2+a^2b+b^2c+c^2a\)
\(\ge2a^2b+2b^2c+2c^2a+a^2b+b^2c+c^2a\)
\(\Rightarrow a^2b+b^2c+c^2a\le\dfrac{1}{3}\left(a+b+c\right)\left(a^2+b^2+c^2\right)=\dfrac{1}{3}\left(a+b+c\right)\)
Lại có:
\(\dfrac{1}{\sqrt{a^2+1}}+\dfrac{1}{\sqrt{b^2+1}}+\dfrac{1}{\sqrt{c^2+1}}\le\sqrt{3\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\right)}\)
\(\Rightarrow VT\le\dfrac{1}{\sqrt{3}}\left(a+b+c\right)\sqrt{\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}}\)
Do đó ta chỉ cần c/m:
\(\dfrac{1}{\sqrt{3}}\left(a+b+c\right)\sqrt{\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}}\le\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\le\dfrac{27}{4\left(a+b+c\right)^2}\)
\(\Leftrightarrow\dfrac{27}{4\left(a+b+c\right)^2}+\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}\ge3\)
Do \(\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3}=\dfrac{\left(a+b+c\right)^2}{4}\)
Nên ta chỉ cần c/m:
\(\dfrac{27}{4\left(a+b+c\right)^2}+\dfrac{\left(a+b+c\right)^2}{4}\ge3\)
\(\Leftrightarrow\dfrac{9}{2\left(a+b+c\right)^2}+\left[\dfrac{9}{4\left(a+b+c\right)^2}+\dfrac{\left(a+b+c\right)^2}{4}\right]\ge3\)
\(\Rightarrow...\)
